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tigry1 [53]
3 years ago
5

G use the margin of error, confidence level, and standard deviation σ to find the minimum sample size required to estimate an un

known population mean μ. margin of error: $121, confidence level: 95%, σ = $528
Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0
The sample size should be 73.

We use the formula

n=(\frac{z*\sigma}{E})^2

To find the z-score:
Convert 95% to a decimal:  0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value goes with a z-score of 1.96.  Using this, our value for σ and our margin of error, we have:
n=(\frac{1.96(528)}{121})^2=73.15\approx73
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1/10, 3/20, 5/30, 7/40 arithmetic sequence geometric sequence neither
adell [148]
If a, b, c is an arithmetic sequence then a + c = 2b

a = 1/10; b = 3/20, c = 5/30

a + c = 1/10 + 5/30 = 3/30 + 5/30 = (3+5)/30 = 8/30
2b = 2 · 3/20 = 3/10 = 9/30

8/30 ≠ 9/30

If a, b, c is a geometric sequence then ac = b²

ac = 1/10 · 5/30 = 1/2 · 1/30 = 1/60
b² = (3/20)² = 9/400

1/60 ≠ 9/400

<span>Answer: Neither.</span>
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3 years ago
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Answer:

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9. m<YXM = 49°

10. m<XWZ = 131°

Step-by-step explanation:

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The perspective is off because <NMW looks to be an obtuse angle because it appears to be greater than a square angle or 90°.

The median should be accurate in perspective even though WZ appears to be greater than MN or XY which is greater than WZ but appears to be smaller than WZ.

6 0
3 years ago
A bucket holds 6 liters of water. If a ladle can scoop out 1/5 of a liter, how many scoops will it take to empty the bucket?
Anna71 [15]

Answer:

The answer is 30 scoops.

Step-by-step explanation:

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3 0
3 years ago
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Tonga ways 150lb. He is loading a freight elevator with identical 72-pound boxes. The elevator can carry no more than 2,000 lb.
Alexxx [7]

Answer:

25 boxes can be loaded.

Step-by-step explanation:

Weight of Tonga = 150 lb

Weight of each box = 72 lb

Maximum weight the elevator can withstand = 2000 lb

Tonga rides along with the boxes.

Let the number of boxes be 'n'.

1 box weight = 72 lb

∴ 'n' boxes weight = 1 box weight × Number of boxes

∴ 'n' boxes weight = 72n

Now, combined weight of boxes and Tonga = 72n+150

Now, the total weight should not exceed the maximum capacity of the elevator.

∴ Combined weight ≤ Maximum Weight

⇒ 72n+150\leq 2000

Now, solving for 'n', we get:

72n\leq 2000-150\\\\72n\leq 1850\\\\n\leq \frac{1850}{72}\\\\n\leq 25.7

So, the number of boxes should be less than or equal to 25.

Therefore, the maximum number of boxes that can be loaded are 25.

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3 years ago
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Answer:

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Step-by-step explanation:

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